Question

Use the convolution integral to find the inverse transforms of:

$\frac{\mathbf{1}}{\left(p+a\right){\left(p+b\right)}^{\mathbf{2}}}$

Short answer

The inverse transform of given equation is $\frac{1}{{\left(b-a\right)}^2}\left[e^{-at}-e^{-bt}-\left(b-a\right)t{e}^{-bt}\right].$

Step-by-step solution

Given information.

The equation is$\frac{1}{\left(p+a\right){\left(p+b\right)}^2}$ .

Inverse transform and Convolution theorem.

The piecewise-continuous and exponentially-restricted real function f(t) is the inverse Laplace transform of a function $\mathit{F}\left(s\right)$, and it has the property:

$$\begin{gathered} \mathit{L}\left\{f\right\}\left(s\right)\mathbf{=}\mathit{L}\left\{f\left(t\right)\right\}\left(s\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{F}\left(s\right) \end{gathered}$$

where L is the Laplace transform.

As per Convolution theorem, if we have two functions, taking their convolution and then Laplace is the same as taking the Laplace first (of the two functions separately) and then multiplying the two Laplace Transforms.

Find the inverse transform of 1(p+a)(p+b)2

Consider the equation.

$\frac{1}{\left(\mathrm{p}+\mathrm{a}\right){\left(\mathrm{p}+\mathrm{b}\right)}^2}=\frac{\mathrm{p}}{\left(\mathrm{p}+\mathrm{a}\right)\left(\mathrm{p}+\mathrm{b}\right)}.\frac{1}{\left(\mathrm{p}+\mathrm{b}\right)}$

As per the convolution theorem.

$$\begin{gathered} L^{-1}\left[\frac{p}{\left(p+a\right)\left(p+b\right)}.\frac{1}{\left(p+b\right)}\right]=L^{-1}\left[G\left(p\right).H\left(p\right)\right] \\ =g*h={\int }_0^{t}g\left(t-\tau \right).h\left(t\right)d\tau \\ ={\int }_0^t\frac{e^{-a\tau }-e^{-b\tau }}{b-a}.e^{-b\left(t-\tau \right)}d\tau \end{gathered}$$

Further solve,

role="math" localid="1659253683865" $$\begin{gathered} L^{-1}\left[\frac{p}{\left(p+a\right)\left(p+b\right)}.\frac{1}{\left(p+b\right)}\right]=\frac{e^{-bt}}{b-a}{\int }_0^t\left(e^{\left(b-a\right)\tau }\right)d\tau \\ =\frac{e^{-bt}}{b-a}{\left[\frac{e^{\left(b-a\right)\tau }}{b-a}-\tau \right]}_0^t \\ =\frac{1}{{\left(b-a\right)}^2}\left[e^{-at}-e^{-bt}\right]-\frac{t{e}^{-bt}}{b-a} \\ =\frac{1}{{\left(b-a\right)}^2}\left[e^{-at}-e^{-bt}-\left(b-a\right)t{e}^{-bt}\right] \end{gathered}$$

Thus, the inverse transform of given equation is $=\frac{1}{{\left(b-a\right)}^2}\left[e^{-at}-e^{-bt}-\left(b-a\right)t{e}^{-bt}\right]$.